Induction Motor Design (3-Phase)



[pic]Induction Motor Design (3-Phase)

Output Equation: - It gives the relationship between electrical rating and physical dimensions (Quantities)

Output of a 3 phase IM is

[pic]

Where

VPh1= Input phase voltage

IPh1= Input Phase current

[pic]

[pic]

Or equation (1) can be written as

[pic]

[pic]

Where

f = frequency of supply =PN/120

P =No of Poles

N =Speed in RPM

Kpd1= Winding factor =0.955

[pic]

[pic]= Average value of fundamental flux density

[pic]=Pole pitch =[pic]

D = Inner diameter of stator

L = Length of the IM

[pic]

[pic]

Total Ampere conductors is known as total electric loading

Specific electric loading

It is defined as electric loading per meter of periphery, denoted by[pic].

[pic]

Or [pic]

Putting the values of f, [pic] & NPh1IPh1 in equation 2 we get

[pic]

[pic]

Or [pic]

Where

[pic]

Choice of magnetic loading ([pic]):

([pic]is average value of fundamental flux density in the air gap)

1. Magnetizing current : [pic]

1. P.F : [pic]

1. Iron Loss : [pic]

1. Heating & Temp rise : [pic]

1. Overload Capacity : [pic]

We know

[pic]

If voltage is constant so for [pic], Nph1 will be less.

And we know

Leakage reactance [pic] Leakage Reactance[pic]

[pic] Isc is more [pic]Dia of circle diagram [pic][pic] Overload Capacity [pic]

6. Noise & Vibration : [pic]

6. Size : [pic]

6. Cost : [pic]

Range of [pic] = 0.3 to 0.6 Tesla

Choice of specific electric loading:

1. Copper Losses : [pic]

1. Heating & Temp Rise : [pic]

1. Overload Capacity : [pic]

If [pic] [pic][pic] NPh1 [pic]

And we know

Leakage reactance [pic]

[pic] Isc is more [pic]Dia of circle diagram [pic][pic] Overload Capacity [pic]

4. Size : [pic]

4. Cost : [pic]

Suitable values of [pic] are

[pic]=10,000 to 17,500 Amp Cond/meter up to 10 KW

=20,000 to 30,000 Amp Cond/meter up to 100 KW

=30,000 to 45,000 Amp Cond/meter > 100 KW

Mini and Maxi value of C:

We know

[pic]

[pic][pic]

[pic]

Effect of Speed on cost and size of IM:

[pic]

So for higher speed IM volume is inversely proportional to speed.

Hence High speed means less volume that is low cost

Estimation of main dimensions (D, L):

We know

[pic]

[pic]

Solving equation (1) & (2) we can find out D & L.

Length of Air gap:

[pic]

Note: D & L are in Meters

[pic]

For medium rating machines

[pic]

Effective length of machine:

Generally

l1= l2= l3=………. = ln

Let

nv =No of ventilating ducts

bv = Width of one ventilating duct

(Generally for every 10 cm of core length there used to be 1 cm ventilating duct)

Gross Iron length

l = l1+ l2+ l3+………. +ln

Actual Iron length

li =Ki*l

Where Ki =Stacking factor

=0.90 to 0.92

Overall length

L = l + nv*bv

Effective length

[pic]

Where [pic] =Effective width of ventilating duct (< bv due to fringing)

Design of Stator:

(1) Shapes of stator slots

May be (i) Open Slot

(ii) Partially Closed Slot

(2) No of Stator Slots S1:

Slot pitch

[pic]

So [pic]

For 3-Phase IM having P-poles

[pic]

Where [pic]

Winding may be integral (q1 is integer) or fractional (q1 is fractional) slot winding.

If q1 is fractional, say

[pic]

Then for windings to be symmetrical it is essential that the denominator ‘y’ should be such that the no of pole pair is divisible by ‘y’.

If double layer winding is to be use then ‘y’ should be divisible by 2.

Hence S1 is estimated.

(3) Estimation of total No of turns per Phase (Nph1), Total no of conductors (Z1) & No of conductors per slots (Nc1):

We know

[pic]

So [pic]

Where

[pic]

[pic]

[pic]

[pic] must be an integer and divisible by 2 for double layer windings. If not an integer make it integer

and hence find the corrected value of [pic] that is [pic]. Also find out the corrected values of

Followings

[pic] Using equation (5)

[pic] Using equation (4)

[pic] Using equation (1)

[pic] Using equation (3)

(4) Sectional area of stator conductor (Fc1):

Per phase stator current

[pic]

So [pic]

Where [pic]

From ICC (Indian Cable Company) table, find dc corresponding to Fc1

SWG | Fc1 (mm) | dc (mm) | doverall (mm)

50 | | 0.025 |

25 | | 0.5 |

1 | | 7.62 |

(5) Stator slot design:

Let

nv = No of conductors vertically

nh = No of conductors horizontally

So Nc1= nv*nh ------------------------- (1)

[pic]

Solving equation (1) & (2) find out nv & nh.

Height of slot

hs1 = nv*dc + 3*0.5 + 3.5 + 1.5 + 2 mm (0.5 mm is insulation thickness and                          2 mm for slack & tolerance)

Width of the slot

bs1 = nh*dc + 2*0.5 + 2 mm (0.5 mm is insulation thickness and                                      2 mm for slack & tolerance)

Slot opening

[pic]

Ratio

[pic]

Thickness of insulation

With mica or leatheroid insulation for small rating machines

KV 0.4 1.1 3.3 6.6 11 15

mm 0.5 0.75 1.5 2.5 4 5.5

With improved insulation (Semica Therm)

KV 2 3 6 10 16 25

mm 1.1 1.4 1.8 2.8 4.0 6.0

Thickness = 0.215 KV +0.7 mm

Advantages of Semica Therm:

(a) Much better heat is dissipated for higher rating machines due to less thickness of wall insulations.

(b) Insulation occupies little less space in the slot.

(6) Length of mean turns (Lmt1)

[pic]

Where [pic]

(7) Resistance of stator winding per phase (RPh1)

[pic]

(8) Total copper loss in the stator winding

[pic]

(9) Flux density in stator tooth

Maximum flux density in stator tooth should not exceed 1.8T; otherwise iron losses and magnetizing current will be abnormally high. (So if flux density >1.8T, change slot dimensions)

Mean flux density in the stator tooth is calculated at [pic] of tooth height from the narrow end of the stator tooth.

Dia of stator at [pic]of tooth height from narrow end

[pic]

Slot pitch at [pic]of tooth height from narrow end

[pic]

Width of the tooth at [pic]of tooth height from narrow end

[pic]

Area of one stator tooth at [pic]of tooth height from narrow end

[pic] (Where li = ki l=Actual iron length)

Area of all the stator teeth under one pole

[pic]

[pic]

[pic]

So mean flux density in teeth

[pic]

(10) Depth (Height) of stator yoke (hy):

Flux through stator yoke is half of the flux per pole

[pic]

Where By = flux density in yoke

= 1.3 to 1.5 T

So [pic]

(11) Outer dia of IM (Do):

[pic]

(12) Estimation of iron losses:

Corresponding to flux density in tooth [pic] find out iron loss per Kg from the graph given on page 19, fig 18. [pic]

Iron loss in teeth

= pit* density * volume of iron in teeth

= pit* 7600 * volume of iron in teeth

Corresponding to flux density in yoke [pic] find out iron loss per Kg from the graph given on page 19, fig 18. [pic]

Iron loss in yoke

= piy* density * volume of iron in yoke

= piy* 7600 * volume of iron in yoke

Total iron losses Pi = Iron loss in teeth + Iron loss in yoke

So [pic]

Rotor Design:

(1) Estimation of rotor no of slots (S2)

If S1= S2, cogging will take place and slot selection also affects noise & vibrations. So as a general rule to avoid crawling, cogging and keeping noise & vibrations low, following slot combinations are selected

[pic]

[pic]

Where, q1 & q2 are no of slots per pole per phase for stator and rotor respectively.

So No of rotor slots

[pic]

(2) Estimation of rotor no of turns, conductors etc

a) Wound rotor IM:

We may keep

[pic]

So No of turns per phase on rotor

[pic]

Total no of conductors on rotor

[pic]

Conductors per slot for rotor

[pic] Make it (NC2) integer if not and divisible by 2 for 2 layer winding. Hence find out correct value of NC2, NPh2 & Z2 i.e. NC2,Corrected NPh2,Corrected ¸ Z2,Corrected

b) Cage rotor IM:

No of rotor bars

[pic][pic]

(3) Rotor current (IPh2)

It is assumed that 85% of ampere turns get transferred to the rotor.

Ampere turns on stator [pic][pic]

a) Wound rotor IM:

Ampere turns on rotor [pic]

So [pic]

Or [pic]

b) Cage rotor IM:

Ampere turns on rotor [pic]

So [pic]

Or [pic]

End ring current

[pic]

(4) Size of rotor conductors:

a) Wound rotor IM:

X-sectional area of rotor conductor

[pic]

Where [pic] = Current density in rotor winding

= 4 to 5 A/mm2

(Higher than stator current density because rotor is rotating so cooling is increased hence, [pic] is more)

SWG or strip conductors may be used.

(b) Cage rotor IM:

X-sectional area of rotor bar

[pic]

Where [pic]= Current density in rotor bar

= 5 to 7 A/mm2

(Higher than stator & wound rotor because rotor conductors are     bare that is no insulation so better heat conduction resulting in     better cooling so [pic] is more)

If round bars are used then dia of bar

[pic]

Or [pic]

X-sectional area of rotor endring

[pic]

Current density in end ring is same as current density in bar.

(5) Flux density in rotor tooth

(Note: This is same as flux density in stator tooth)

Dia of rotor at [pic]of tooth height from narrow end

[pic]

Slot pitch at [pic]of tooth height from narrow end

[pic]

Width of the tooth at [pic]of tooth height from narrow end

[pic]

Area of one stator tooth at [pic]of tooth height from narrow end

[pic]

Area of all the stator teeth under one pole

[pic]

[pic]

[pic]

So mean flux density in teeth

[pic]

(6) Rotor copper loss

a) Wound rotor:

Length of mean turns of rotor

[pic]

DC resistance per phase at 75 0 C

[pic]

We don’t take the ac resistance because the rotor current frequency is very small (f2=sf)

So Rotor cu loss [pic]

b) Cage rotor:

Resistance of one bar [pic]

Cu loss in bars [pic]

Resistance of end ring

[pic]

Cu loss in end rings

[pic]

Total cu loss = Cu loss in bars + Cu loss in end rings

Slip [pic]

[pic]

Losses = Rotor Iron Loss (Negligible) + Rotor Cu loss + F & W loss

F & W loss up to 5% for small motors

3% to 4% for medium motors

2% to 3% for large motors

S up to 5% for small motors

2.5% to 3.5% for medium motors

1% to 1.5% for large motors

Effective air gap length[pic]:

Effective air gap length

[pic]

Where [pic]

[pic]

[pic]

[pic]

K01 & K02 are constants

MMF required in air GAP[pic]:

[pic] AT/m

[pic]

Flux density distribution:

Flux density at 300 from direct axis

= flux density at 600 from inter-polar axis

So [pic]

[pic]

[pic]

For all practical purposes this value is modified to

[pic]

Estimation of magnetizing current & No load current (Im & Io):

|S.No |Part |Length of path |Flux density |at (AT/m) |ATpole-pair |

|1 |Stator Yoke |ly |By |aty |ATy |

|2 |Stator Tooth |2ht1 |[pic] |at2ht1 |AT2ht1 |

|3 |Air Gap |[pic] |[pic] |[pic] |[pic] |

|4 |Rotor Tooth |2ht2 |[pic] |at2ht2 |AT2ht2 |

|5 |Rotor Yoke |lry |Bry |atry |ATry |

|ATpole-pair =[pic]= [pic] |

[pic]

AT for one pole [pic]

So no load current

[pic] [pic]

No load current

[pic]

No load power factor

[pic]

Estimations of Ideal Blocked rotor current:

Total resistance referred to stator

[pic] [pic]

Total leakage reactance referred to stator

[pic] [pic]

[pic]

[pic]

Estimation of leakage reactance:

Leakage reactance consists of

1. Stator slot leakage reactance

1. Rotor slot leakage reactance

(a) Wound rotor or

(b) Cage rotor

3. Overhang or end turns leakage reactance

4. Zigzag leakage reactance

For cage rotor IM Zigzag leakage reactance is small and may be ignored.

5. Differential or harmonic leakage reactance

1. Stator slot leakage reactance:

Assumptions are

(i) Permeability of iron is infinity so NO MMF is consumed in iron path.

(ii) Leakage flux path is parallel to slot width

Let

Ic1 = Conductor current (A)

Zc1 = No of conductors per slot

Z1 = Total No of conductors

NPh1 = Turns per Phase

P = No of poles

q1 = Slot / Pole /Phase

(a) For 1-Layer winding

Total amp conductors in slot =Ic1 Zc1

Consider an elementary path of thickness dx at a distance of x as shown in the figure. Let [pic]be the leakage flux through the elementary path of thickness dx & height x.

MMF at distance x

[pic] ------ (1)

Permeance [pic] ----- (2)

So

[pic]

[pic] ---- (3)

Leakage flux linkages associated with this elementary path

[pic]

[pic] ---------- (4)

So flux linkages in height h1

[pic] ------------------- (5) (Integrating equation 4 from 0 to h1)

[pic] ------------------- (6)

Leakage flux linkages in height h2

[pic] ------------------- (7)

Leakage flux linkages in height h3

[pic] ------------------- (8)

Leakage flux linkages in height h4

[pic] ------------------- (9)

Total slot leakage flux

[pic]----------- (10)

Slot leakage inductance will be

[pic]--------- (11)

[pic] --------- (12)

Where

[pic]

No of slots per phase =Pq1

Slot leakage inductance per phase

[pic] ---------- (13)

Total No of conductors

[pic]

So [pic] Put in above equation

So [pic]

Or [pic] -------- (14)

Slot leakage reactance per phase (1-Layer)

[pic]

[pic] -------- (15)

(b) For 2-Layer winding

Same as 1-Layer winding

Slot leakage reactance per phase (2-layer)

[pic] -------- (16)

2. Rotor slot leakage reactance(X2)

(a) Wound rotor: Estimated in the same manor as for stator.

(b) Cage rotor:

W0 = b02 = (0.2 to 0.4) d2bar

h = 1 to 3 mm

Rotor reactance per phase

[pic] ----------------- (17)

Where

[pic]

[pic]

Rotor resistance referred to stator

[pic]

Where

[pic]

3. Overhang leakage reactance(X0):

[pic] -------------- (18)

Where

[pic]

[pic]

l0 = Length of conductor in overhang

Ks = Slot leakage factor

[pic]

[pic]

4. Zigzag leakage reactance (Xz):

[pic] --------------- (19)

Where

[pic] [pic] & [pic]

5. Differential or Belt or Harmonic leakage reactance (Xh):

It is ignored for cage rotor but considered for wound rotor IM. It is due to the fact that spatial distribution of MMFs of the primary and secondary windings is not the same; the difference in the harmonic contents of the two MMFs causes harmonic leakage fluxes.

[pic] ----------------- (20)

Where

Kh1 & Kh2 are the factors for stator & rotor

Hence

Total leakage reactance referred to the stator side

[pic]

Construction of Circle Diagram from designed data:

We should know following for drawing the circle diagram

a. No load current and no load power factor

a. Short circuit current and short circuit power factor

Steps to draw the circle diagram are (See the figure on next page)

1. Draw horizontal (x-axis) and vertical (y-axis) lines.

1. Draw I0 at an angle [pic] from vertical line assuming some scale for current.

1. Draw Isc at an angle [pic] from vertical line.

1. Join AB, which represents the o/p line of the motor to power scale.

1. Draw a horizontal line AF, and erect a perpendicular bisector on the o/p line AB so as to meet the line AF at the point O’. Then O’ as center and AO’ as radius, draw a semi circle ABF.

1. Draw vertical line BD; divide line BD in the radio of rotor copper loss to stator copper loss at the point E.

1. Join AE, which represent the torque line.

Determination of design performance from above circle diagram

1. Power scale can be find out from current scale

Power in watt per Cm = Voltage x Current per Cm

2. Full load current & power factor

Draw a vertical line BC representing the rated o/p of the motor s per the power scale. From point C, draw a line parallel to o/p line, so as to cut the circle at pint P. Join OP which represents the full load current of the motor to current scale. Operating power factor can also be found out.

3. Full load efficiency

Draw a vertical line from P as shown in above figure.

PL = O/p Power

PX = I/p Power

[pic]

4. Full load slip

[pic]

[pic]

5. Starting torque

Line BE represents the starting torque of the motor in synchronous watts to power scale

We can also draw maximum torque and maximum power output from circle diagram.

(Not shown in above diagram)

-----------------------

L

ln

l3

l2

l1

bv

Partially closed slot for 400Volts IM

hs1

Wedge

Tooth lip

3.5 mm

b01

1.5 mm

bs1

One Turns

L

Coil Span

< Pole pitch

hy

Teeth

Slot

Yoke

2d2bar

d2bar

[pic]

[pic]

[pic]

[pic]

Rotor

[pic]

[pic]

Stator

Zigzag leakage Reactance

Rotor

Stator

Zigzag leakage flux

Over Hang Leakage Reactance

L

Over Hang Leakage flux

Slot leakage Reactance

Leakage flux

MMF Distribution

dx

x

Stator slot

h1

h3

b01

h4

bs1

h2

Zc1Ic1

Mx

h1

h1

MMF Distribution

2-Layer stator slot

h4

b01

h5

bs1

h3

Zc1Ic1

h2

[pic]

0.5 mm

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